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HCl + Al + Na2MoO4O4 = H2O + NaCl + AlCl3 + MoCl2

Input interpretation

HCl hydrogen chloride + Al aluminum + Na2MoO4O4 ⟶ H_2O water + NaCl sodium chloride + AlCl_3 aluminum chloride + Cl_2Mo_1 molybdenum(II) chloride
HCl hydrogen chloride + Al aluminum + Na2MoO4O4 ⟶ H_2O water + NaCl sodium chloride + AlCl_3 aluminum chloride + Cl_2Mo_1 molybdenum(II) chloride

Balanced equation

Balance the chemical equation algebraically: HCl + Al + Na2MoO4O4 ⟶ H_2O + NaCl + AlCl_3 + Cl_2Mo_1 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 Al + c_3 Na2MoO4O4 ⟶ c_4 H_2O + c_5 NaCl + c_6 AlCl_3 + c_7 Cl_2Mo_1 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, Al, Na, Mo and O: Cl: | c_1 = c_5 + 3 c_6 + 2 c_7 H: | c_1 = 2 c_4 Al: | c_2 = c_6 Na: | 2 c_3 = c_5 Mo: | c_3 = c_7 O: | 8 c_3 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 16 c_2 = 4 c_3 = 1 c_4 = 8 c_5 = 2 c_6 = 4 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HCl + 4 Al + Na2MoO4O4 ⟶ 8 H_2O + 2 NaCl + 4 AlCl_3 + Cl_2Mo_1
Balance the chemical equation algebraically: HCl + Al + Na2MoO4O4 ⟶ H_2O + NaCl + AlCl_3 + Cl_2Mo_1 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HCl + c_2 Al + c_3 Na2MoO4O4 ⟶ c_4 H_2O + c_5 NaCl + c_6 AlCl_3 + c_7 Cl_2Mo_1 Set the number of atoms in the reactants equal to the number of atoms in the products for Cl, H, Al, Na, Mo and O: Cl: | c_1 = c_5 + 3 c_6 + 2 c_7 H: | c_1 = 2 c_4 Al: | c_2 = c_6 Na: | 2 c_3 = c_5 Mo: | c_3 = c_7 O: | 8 c_3 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_3 = 1 and solve the system of equations for the remaining coefficients: c_1 = 16 c_2 = 4 c_3 = 1 c_4 = 8 c_5 = 2 c_6 = 4 c_7 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 16 HCl + 4 Al + Na2MoO4O4 ⟶ 8 H_2O + 2 NaCl + 4 AlCl_3 + Cl_2Mo_1

Structures

+ + Na2MoO4O4 ⟶ + + +
+ + Na2MoO4O4 ⟶ + + +

Names

hydrogen chloride + aluminum + Na2MoO4O4 ⟶ water + sodium chloride + aluminum chloride + molybdenum(II) chloride
hydrogen chloride + aluminum + Na2MoO4O4 ⟶ water + sodium chloride + aluminum chloride + molybdenum(II) chloride

Equilibrium constant

Construct the equilibrium constant, K, expression for: HCl + Al + Na2MoO4O4 ⟶ H_2O + NaCl + AlCl_3 + Cl_2Mo_1 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 16 HCl + 4 Al + Na2MoO4O4 ⟶ 8 H_2O + 2 NaCl + 4 AlCl_3 + Cl_2Mo_1 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HCl | 16 | -16 Al | 4 | -4 Na2MoO4O4 | 1 | -1 H_2O | 8 | 8 NaCl | 2 | 2 AlCl_3 | 4 | 4 Cl_2Mo_1 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 16 | -16 | ([HCl])^(-16) Al | 4 | -4 | ([Al])^(-4) Na2MoO4O4 | 1 | -1 | ([Na2MoO4O4])^(-1) H_2O | 8 | 8 | ([H2O])^8 NaCl | 2 | 2 | ([NaCl])^2 AlCl_3 | 4 | 4 | ([AlCl3])^4 Cl_2Mo_1 | 1 | 1 | [Cl2Mo1] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HCl])^(-16) ([Al])^(-4) ([Na2MoO4O4])^(-1) ([H2O])^8 ([NaCl])^2 ([AlCl3])^4 [Cl2Mo1] = (([H2O])^8 ([NaCl])^2 ([AlCl3])^4 [Cl2Mo1])/(([HCl])^16 ([Al])^4 [Na2MoO4O4])
Construct the equilibrium constant, K, expression for: HCl + Al + Na2MoO4O4 ⟶ H_2O + NaCl + AlCl_3 + Cl_2Mo_1 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 16 HCl + 4 Al + Na2MoO4O4 ⟶ 8 H_2O + 2 NaCl + 4 AlCl_3 + Cl_2Mo_1 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HCl | 16 | -16 Al | 4 | -4 Na2MoO4O4 | 1 | -1 H_2O | 8 | 8 NaCl | 2 | 2 AlCl_3 | 4 | 4 Cl_2Mo_1 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HCl | 16 | -16 | ([HCl])^(-16) Al | 4 | -4 | ([Al])^(-4) Na2MoO4O4 | 1 | -1 | ([Na2MoO4O4])^(-1) H_2O | 8 | 8 | ([H2O])^8 NaCl | 2 | 2 | ([NaCl])^2 AlCl_3 | 4 | 4 | ([AlCl3])^4 Cl_2Mo_1 | 1 | 1 | [Cl2Mo1] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HCl])^(-16) ([Al])^(-4) ([Na2MoO4O4])^(-1) ([H2O])^8 ([NaCl])^2 ([AlCl3])^4 [Cl2Mo1] = (([H2O])^8 ([NaCl])^2 ([AlCl3])^4 [Cl2Mo1])/(([HCl])^16 ([Al])^4 [Na2MoO4O4])

Rate of reaction

Construct the rate of reaction expression for: HCl + Al + Na2MoO4O4 ⟶ H_2O + NaCl + AlCl_3 + Cl_2Mo_1 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 16 HCl + 4 Al + Na2MoO4O4 ⟶ 8 H_2O + 2 NaCl + 4 AlCl_3 + Cl_2Mo_1 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HCl | 16 | -16 Al | 4 | -4 Na2MoO4O4 | 1 | -1 H_2O | 8 | 8 NaCl | 2 | 2 AlCl_3 | 4 | 4 Cl_2Mo_1 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HCl | 16 | -16 | -1/16 (Δ[HCl])/(Δt) Al | 4 | -4 | -1/4 (Δ[Al])/(Δt) Na2MoO4O4 | 1 | -1 | -(Δ[Na2MoO4O4])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) NaCl | 2 | 2 | 1/2 (Δ[NaCl])/(Δt) AlCl_3 | 4 | 4 | 1/4 (Δ[AlCl3])/(Δt) Cl_2Mo_1 | 1 | 1 | (Δ[Cl2Mo1])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/16 (Δ[HCl])/(Δt) = -1/4 (Δ[Al])/(Δt) = -(Δ[Na2MoO4O4])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/2 (Δ[NaCl])/(Δt) = 1/4 (Δ[AlCl3])/(Δt) = (Δ[Cl2Mo1])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)
Construct the rate of reaction expression for: HCl + Al + Na2MoO4O4 ⟶ H_2O + NaCl + AlCl_3 + Cl_2Mo_1 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 16 HCl + 4 Al + Na2MoO4O4 ⟶ 8 H_2O + 2 NaCl + 4 AlCl_3 + Cl_2Mo_1 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HCl | 16 | -16 Al | 4 | -4 Na2MoO4O4 | 1 | -1 H_2O | 8 | 8 NaCl | 2 | 2 AlCl_3 | 4 | 4 Cl_2Mo_1 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HCl | 16 | -16 | -1/16 (Δ[HCl])/(Δt) Al | 4 | -4 | -1/4 (Δ[Al])/(Δt) Na2MoO4O4 | 1 | -1 | -(Δ[Na2MoO4O4])/(Δt) H_2O | 8 | 8 | 1/8 (Δ[H2O])/(Δt) NaCl | 2 | 2 | 1/2 (Δ[NaCl])/(Δt) AlCl_3 | 4 | 4 | 1/4 (Δ[AlCl3])/(Δt) Cl_2Mo_1 | 1 | 1 | (Δ[Cl2Mo1])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/16 (Δ[HCl])/(Δt) = -1/4 (Δ[Al])/(Δt) = -(Δ[Na2MoO4O4])/(Δt) = 1/8 (Δ[H2O])/(Δt) = 1/2 (Δ[NaCl])/(Δt) = 1/4 (Δ[AlCl3])/(Δt) = (Δ[Cl2Mo1])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

Chemical names and formulas

| hydrogen chloride | aluminum | Na2MoO4O4 | water | sodium chloride | aluminum chloride | molybdenum(II) chloride formula | HCl | Al | Na2MoO4O4 | H_2O | NaCl | AlCl_3 | Cl_2Mo_1 Hill formula | ClH | Al | MoNa2O8 | H_2O | ClNa | AlCl_3 | Cl_2Mo name | hydrogen chloride | aluminum | | water | sodium chloride | aluminum chloride | molybdenum(II) chloride IUPAC name | hydrogen chloride | aluminum | | water | sodium chloride | trichloroalumane | dichloromolybdenum
| hydrogen chloride | aluminum | Na2MoO4O4 | water | sodium chloride | aluminum chloride | molybdenum(II) chloride formula | HCl | Al | Na2MoO4O4 | H_2O | NaCl | AlCl_3 | Cl_2Mo_1 Hill formula | ClH | Al | MoNa2O8 | H_2O | ClNa | AlCl_3 | Cl_2Mo name | hydrogen chloride | aluminum | | water | sodium chloride | aluminum chloride | molybdenum(II) chloride IUPAC name | hydrogen chloride | aluminum | | water | sodium chloride | trichloroalumane | dichloromolybdenum

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